HSC Extension 1 and 2 Mathematics/3-Unit/HSC/Primitive of sin and cos squared

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The integration of these functions can be performed by the substitution of \sin^2\alpha or \cos^2\alpha by a function in terms of \cos2\alpha respectively.

Deriving the substituted functions[edit]

By the double-angle formula, \left.\cos 2\alpha = 2\cos^2 \alpha - 1\right.. Rearranging to make \cos^2\alpha the subject, \cos^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha.

We know by the Pythagorean identity \left.\sin^2\theta + \cos^2\theta = 1\right. that \left.\cos^2\alpha = 1 - \sin^2\alpha\right., so 1 - \sin^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha. Rearranging, \sin^2\alpha = \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha.

Finding the primitive[edit]

By substituting \sin^2\alpha\, for \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha, \int \sin^2 \alpha d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha d\alpha. Similarly for \cos^2\alpha \,,

\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha

These can be integrated by using the primitives of cosine and sine

 \int \cos \alpha d\alpha = \sin \alpha + c

 \int \sin \alpha d\alpha = - \cos \alpha + c

For cos2 α

\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha

\int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c

\int \cos^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c

For sin2 α


\int \sin^2 \alpha \, d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha

\int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c

\int \sin^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c